Question
Verify the following identity \dfrac{\sin 4x+\sin 6x}{\sin 4x-\sin 6x} =-\tan 5x\cot x
Answer
Apply the sum-to-product identity for sin\alpha +\sin\beta to the numerator
\sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}
Then,
\sin 4x+\sin 6x =2\sin\dfrac{4x+6x}{2}\cos\dfrac{4x-6x}{2}
\sin 4x+\sin 6x =2\sin 5x\cos(-x)
\because \cos(-x) =\cos x
\therefore \sin 4x+\sin 6x =2\sin 5x\cos x
(1)
Apply the sum-to-product identity for sin\alpha -\sin\beta to the denominator
\sin \alpha - \sin \beta = 2\cos\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}
Then,
\sin 4x-\sin 6x =2\cos\dfrac{4x+6x}{2}\sin\dfrac{4x-6x}{2}
\sin 4x-\sin 6x = 2\cos 5x\sin(-x)
\because \sin(-x)=-\sin x
\therefore \sin 4x-\sin 6x = -2\cos 5x\sin x
(2)
Substitute (1) and (2) into the LHS of the identity
\dfrac{\sin 4x+\sin 6x}{\sin 4x-\sin 6x}
=\dfrac{2\sin 5x\cos x}{-2\cos 5x\sin x}
=-\tan 5x\cot x
Now we have verified the identity
\dfrac{\sin 4x+\sin 6x}{\sin 4x-\sin 6x} =-\tan 5x\cot x